




height h=2r 



acceleration due to gravity g 



Height x less than or equal to h , x 



Put any k value 







FIND 



velocity at B so that the particle does not fall off at A:
u=√(5gr) 



vx=√(2gx) 



velocity at B so that the particle at A is zero: u1=√(2gr) 



velocity at A when velocity at B is u v=√(gr) 



velocity at A when velocity at B is u1v1=0 



radius of the circle r: 



velocity at P corresponding to u, vp=√(u^{2}2gx) 



velocity at P corresponding to u1, vp1=√(u1^{2}2gx) 



Angle between OP and OB :θ in degree 



OX 



XP 



Path length BP 



Displacement BP 



Path length/displacement 



sinθ 



cosθ 



tanθ 



Time period of revolution=T 



threshhold Time period of revolution=T1 i.e particle shall reach
the top and then free fall to ground. 



Frequency f : 



x^{2} = vx^{2}/u^{2} ; 



y^{2} = vp^{2}/u^{2} ; 



x^{2}+y^{2} =1 



x 



y 



Let velocity at B be u2=√(5+x)gr 



The trajectory shall be a circle of radius r2=(5+x)r/5=(1+0.2*k)r 



velocity at A when velocity at B is u2is v2=√(gr2) 



When ground velocity is changed , the radius of the circle
changes so that the apex velocity shall remain √(gr') and ground
velocity is √(5gr'). When we revolve an object attached to a
string, with increasing velocity, speed also increases at every
point and the corresponding force is balanced by string tension till
the string snaps. 

















